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Câu 1: Sửa lạ đề chút nhé : 4x + 1 -> 4x -1
Đặt A = \(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)
=> \(\sqrt{2}.A\)= \(\sqrt{4x-1+2\sqrt{4x-1}+1}+\sqrt{4x-1-2\sqrt{4x-1}+1}\)
= \(\sqrt{\left(\sqrt{4x-1}+1\right)^2}+\sqrt{\left(\sqrt{4x-1}-1\right)^2}\)
= \(\left|\sqrt{4x-1}+1\right|+\left|\sqrt{4x-1}-1\right|\)
Vì \(\frac{1}{4}< x< \frac{1}{2}\Rightarrow0< 4x-1< 1\Rightarrow0< \sqrt{4x-1}< 1\)
nên \(\sqrt{2}A=\)\(\sqrt{4x-1}+1+1-\sqrt{4x-1}\)=2
=> \(A=2:\sqrt{2}=\sqrt{2}\)
Câu 2. Có: \(9-4\sqrt{2}=8-2.2\sqrt{2}+1=\left(2\sqrt{2}-1\right)^2\)
=> \(\sqrt{9-4\sqrt{2}}=2\sqrt{2}-1\)
=> \(4+\sqrt{9-4\sqrt{2}}=4+2\sqrt{2}-1=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)
=> \(\sqrt{4+\sqrt{9-4\sqrt{2}}}=\sqrt{2}+1\)
=> \(53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}=53-20\left(\sqrt{2}+1\right)=33-2.10\sqrt{2}=5^2-2.5.2\sqrt{2}+8=\left(5-2\sqrt{2}\right)^2\)
=> \(\sqrt{53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}}=5-2\sqrt{2}\)
\(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)
\(A=3\sqrt{8}-\sqrt{50}-\sqrt{\sqrt{2}-1}\)
\(\Leftrightarrow6\sqrt{2}-5\sqrt{2}-\sqrt{\sqrt{2}-1}\)
\(\Leftrightarrow\sqrt{2}-\sqrt{\sqrt{2}-1}\)
\(B=2.\dfrac{2}{x-1}.\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{\sqrt{x^2-2x+1}}{2x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{\sqrt{\left(x-1\right)^2}}{x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{x-1}{x}\)
\(\Leftrightarrow\)\(2.\dfrac{1}{x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x}\)
a)\(\)https://www.cymath.com/answer?q=2sqrt(27)-6sqrt(4%2F3)%2B3%2F5sqrt(75)
\(M=2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}=2\sqrt{3^2.3}-6\sqrt{\frac{2^2.3}{3^2}}+\frac{3}{5}\sqrt{5^2.3}=.\)
\(=6\sqrt{3}-4\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
\(P=\frac{2}{x-1}\sqrt{\frac{x^2-2x+1}{4x^2}}.Với...0< x< 1\Leftrightarrow\) \(P=\frac{2}{x-1}\sqrt{\frac{\left(x-1\right)^2}{\left(2x\right)^2}}=\frac{2}{(x-1)}.\frac{\left(1-x\right)}{2x}=\frac{-1}{x}.\)
\(M=2\sqrt{3^2.3}-6\frac{\sqrt{2^2.3}}{3}+\frac{3}{5}\sqrt{5^2.3}\)
\(M=6\sqrt{3}-4\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
\(P=\frac{2}{x-1}\sqrt{\frac{\left(x-1\right)^2}{\left(2x\right)^2}}=\frac{2}{x-1}.\frac{\left|x-1\right|}{2x}=\frac{-2\left(x-1\right)}{\left(x-1\right).2x}=-\frac{1}{x}\)
Anh hai nhanh tay hơn em nghĩ đó. Em làm xong rùi, chụp ảnh đang định gửi lên thì thấy tên anh đập ngay vào mắt. Haiz, thất vọng não nề!!
a/ \(\sqrt{2x-3}=\sqrt{x-1}ĐK:x\ge\dfrac{3}{2}\)
\(\Leftrightarrow2x-3=x-1\Leftrightarrow x=-1+3=2\)(tm)
b/ \(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)ĐK: x≥1
\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
\(\Leftrightarrow\sqrt{x-1}\left(6-3-2+1\right)=16\)
\(\Leftrightarrow2\sqrt{x-1}=16\Leftrightarrow\sqrt{x-1}=8\Leftrightarrow x-1=64\Leftrightarrow x=65\)
(tm)
c/ \(\sqrt{2x+3}+\sqrt{2x+2}=1\)ĐK: x>=-1
\(\Leftrightarrow\sqrt{2x+3}=1-\sqrt{2x+2}\)
\(\Leftrightarrow2x+3=2x+2-2\sqrt{2x+2}+1\)
\(\Leftrightarrow2\sqrt{2x+2}=0\Leftrightarrow\sqrt{2x+2}=0\Leftrightarrow2x+2=0\Leftrightarrow x=-1\left(tm\right)\)
d/ \(\sqrt{4x^2+4x+1}=3\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=3\Leftrightarrow\left|2x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy....
a)\(x+3+\sqrt{x^2-6x+9}\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+x-3\)
\(=2x\)
b)\(\sqrt{x^2+4x+4}-\sqrt{x^2}\)
\(=\sqrt{\left(x+2\right)^2}-x\)
\(=x+2-x\)
=2
c)\(\sqrt{\frac{x^2-2x+1}{x-1}}\)
\(=\sqrt{\frac{\left(x-1\right)^2}{x-1}}\)
\(=\sqrt{x-1}\)
\(P=\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\) với \(\dfrac{1}{4}< x< \dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{2}P=\sqrt{4x+2\sqrt{4x-1}}+\sqrt{4x-2\sqrt{4x-1}}\)
\(=\sqrt{\left(\sqrt{4x-1}\right)^2+2\sqrt{4x-1}+1}+\sqrt{\left(\sqrt{4x-1}\right)^2-2\sqrt{4x-1}+1}\)
\(=\sqrt{4x-1}+1+\left|\sqrt{4x-1}-1\right|\)
Do \(\dfrac{1}{4}< x< \dfrac{1}{2}\Leftrightarrow0< \sqrt{4x-1}< 1\)
\(\Rightarrow P=\dfrac{1}{\sqrt{2}}\left(\sqrt{4x-1}+1+1-\sqrt{4x-1}\right)=\sqrt{2}\)
Vậy \(P=\sqrt{2}\).